If $$x^2 - 4x + 4b = 0$$ has two real solutions, find the value of 'b'.

Given,

 $$x^2 - 4x + 4b = 0$$ has two real solutions.

So the discriminant must equal or greater than 0.

On comparing the given equation with $$ax^2 + bx + c = 0$$

a = 1 ; b = - 4 ; c = 4b

substituting the values in equation $$b^2 - 4ac ≥ 0$$,
$$(-4)^2 - 4(1)(4b) ≥ 0$$

$$16 - 16b ≥ 0$$

$$16 ≥ 16b$$

$$1 ≥ b$$

Thus,

The required value of b is either 1 or lesser than 1. As in option only $$b<1$$ is represented, the answer is option D-> $$b<1$$.